NEET · Physics · STD 12 - 4. Moving charges and magnetism
In a uniform magnetic field of \(0.049 \mathrm{~T}\), a magnetic needle performs \(20\) complete oscillations in \(5\) seconds as shown. The moment of inertia of the needle is \(9.8 \times 10^{-5} \mathrm{~kg} \mathrm{~m}^2\). If the magnitude of magnetic moment of the needle is \(x \times 10^{-5} \mathrm{Am}^2\), then the value of ' \(x\) ' is :
- A \(128 \pi^2\)
- B \(50 \pi^2\)
- C \(1280 \pi^2\)
- D \(5 \pi^2\)
Answer & Solution
Correct Answer
(C) \(1280 \pi^2\)
Step-by-step Solution
Detailed explanation
Time period of Oscillation, \(T=2 \pi \sqrt{\frac{1}{M B}}\) \(\Rightarrow \frac{1}{4}=2 \pi \sqrt{\frac{9.8 \times 10^{-6}}{M \times 0.049}}\)
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